1. **State the problem:** Evaluate the integral $$\int \frac{\sin 3x}{1 + \cos 3x} \, dx.$$\n\n2. **Recall the formula and rules:** We can use the substitution method and trigonometric identities to simplify the integral. Note that \(1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}\) and \(\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\).\n\n3. **Rewrite the integral using substitution:** Let \(u = 1 + \cos 3x\), then \(du = -3 \sin 3x \, dx\) or \(\sin 3x \, dx = -\frac{1}{3} du\).\n\n4. **Substitute into the integral:** $$\int \frac{\sin 3x}{1 + \cos 3x} \, dx = \int \frac{\sin 3x \, dx}{u} = \int \frac{-\frac{1}{3} du}{u} = -\frac{1}{3} \int \frac{1}{u} \, du.$$\n\n5. **Integrate:** $$-\frac{1}{3} \int \frac{1}{u} \, du = -\frac{1}{3} \ln |u| + C = -\frac{1}{3} \ln |1 + \cos 3x| + C.$$\n\n**Final answer:** $$\boxed{-\frac{1}{3} \ln |1 + \cos 3x| + C}.$$